Wednesday, April 13, 2016

Math Is Power

1.0 Introduction

A common type of post in this blog is the presentation of concrete numerical examples in economics. Sometimes I present examples to illustrate some principle. But usually I try to find examples that are counter-intuitive or perverse, at least from the perspective of economics as mainstream economists often misteach it.

Voting games provide an arena where one can find surprising results in political science. I am thinking specifically of power indices. In this post, I try to explain two of the most widely used power indices by means of an example.

2.0 Me and My Aunt: A Voting Game

For purposes of exposition, I consider a specific game, called Me and My Aunt. There are four players in this version of the game, represented by elements of the set:

P = The set of players = {0, 1, 2, 3}

Out of respect, the first player gets two votes, while all other players get a vote each (Table 1). A coalition, S, is a set of players. That is, a coalition is a subset of P. A coalition passes a resolution if it has a majority of votes. Since there are four players, one of who has two votes, the total number of votes is five. So a majority, in this game of weighted voting, is three votes.

Table 1: Players and Their Votes

Players

Votes

0 (Aunt)

2

1 (Me)

1

2

1

3

1

One needs to specify the payoff to each coalition to complete the definition, in characteristic function form, of this game. The characteristic function, v(S) maps the set of all subsets of P to the set {0, 1}. If the players in S have three or more votes,v(S) is 1. Otherwise, it is 0. That is, a winning coalition gains a payoff of one to share among its members.

3.0 The Penrose-Banzhaf Power Index

Power for a player, in this mathematical approach, is the ability to be the decisive member of a coalition. If, for a large number of coalitions, you being in or out of a coalition determines whether or not that coalition can pass a resolution, you have a lot of power. Correspondingly, if the members of most coalitions do not care whether you join, because your presence has no influence on whether or not they can put their agenda into effect, you have little power.

The Penrose-Banzhaf power index is one (of many) attempts to quantify this idea. Table 2 lists all 16 coalitions for the voting game under consideration. (The number of coalitions is the sum of a row in Pascal's triangle.) The second column in Table 2 specifies the value for the characteristic function for that coalition. Equivalently, the third column notes which eight coalitions are winning coalitions, and which eight are losing. The last two columns are useful for tallying up counts needed for the Penrose-Banzhaf index.

Table 2: Calculations for Penrose-Banzhaf Power Index

Coalition

CharacteristicFunction

Winningor Losing

Player

Aunt (0)

Me (1)

{}

v( {} ) = 0

Losing

0

0

{0}

v( {0} ) = 0

Losing

0

0

{1}

v( {1} ) = 0

Losing

0

0

{2}

v( {2} ) = 0

Losing

0

0

{3}

v( {3} ) = 0

Losing

0

0

{0, 1}

v( {0, 1} ) = 1

Winning

1

1

{0, 2}

v( {0, 2} ) = 1

Winning

1

0

{0, 3}

v( {0, 3} ) = 1

Winning

1

0

{1, 2}

v( {1, 2} ) = 0

Losing

0

0

{1, 3}

v( {1, 3} ) = 0

Losing

0

0

{2, 3}

v( {2, 3} ) = 0

Losing

0

0

{0, 1, 2}

v( {0, 1, 2} ) = 1

Winning

1

0

{0, 1, 3}

v( {0, 1, 3} ) = 1

Winning

1

0

{0, 2, 3}

v( {0, 2, 3} ) = 1

Winning

1

0

{1, 2, 3}

v( {1, 2, 3} ) = 1

Winning

0

1

{0, 1, 2, 3}

v( {0, 1, 2, 3} ) = 1

Winning

0

0

The Penrose-Banzhaf index, ψ(i) is calculated for each player i. It is defined, for a given player, to be the ratio of the number of winning coalitions in which that player is decisive to the total number of coalitions, winning or losing. A player is decisive for a coalition if:

The coalition is a winning coalition.

The removal of the player from the coalition converts it to a losing coalition.

From the table above, one can see that player 0 is decisive for six coalitions, while player 1 is decisive for only two coalitions. Hence, the Penrose-Banzhaf index for "my aunt" is:

ψ(0) = 6/16 = 3/8

By symmetry, the index values for players 2 and 3 are the same as the value for player 1:

ψ(1) = ψ(2) = ψ(3) = 2/16 = 1/8

More than one player can be decisive for a winning coalition. No need exists for the Penrose-Banzhaf index to sum up to one. How much one's vote is weighted does not bear a simple relationship to how much power one has. Also note that the definition of this power index is not confined to simple majority games. Power indices can be calculated for voting games in which a super-majority is required to pass a measure. For example, in the United States Senate, 60 senators are needed to end a filibuster.

4.0 The Shapley-Shubik Power Index

The Shapley-Shubik power index is an application of the calculation of the Shapley value to voting games. The Shapley value applies to cooperative games, in general. For its use as a measure of power in voting games, it matters in which order players enter a coalition. Accordingly, Table 3 lists all 24 permutations of all four players in the voting game being analyzed.

Table 3: Calculations for the Shapley-Shubik Power Index

Permutation

Player

Aunt (0)

Me (1)

(0, 1, 2, 3)

v( {0} ) - v( {} ) = 0

v( {0, 1} ) - v( {0} ) = 1

(0, 1, 3, 2)

v( {0} ) - v( {} ) = 0

v( {0, 1} ) - v( {0} ) = 1

(0, 2, 1, 3)

v( {0} ) - v( {} ) = 0

v( {0, 1, 2} ) - v( {0, 2} ) = 0

(0, 2, 3, 1)

v( {0} ) - v( {} ) = 0

v( {0, 1, 2, 3} )- v( {0, 2, 3} ) = 0

(0, 3, 1, 2)

v( {0} ) - v( {} ) = 0

v( {0, 1, 3} )- v( {0, 3} ) = 0

(0, 3, 2, 1)

v( {0} ) - v( {} ) = 0

v( {0, 1, 2, 3} )- v( {0, 2, 3} ) = 0

(1, 0, 2, 3)

v( {0, 1} )- v( {1} ) = 1

v( {1} ) - v( {} ) = 0

(1, 0, 3, 2)

v( {0, 1} )- v( {1} ) = 1

v( {1} ) - v( {} ) = 0

(1, 2, 0, 3)

v( {0, 1, 2} )- v( {1, 2} ) = 1

v( {1} ) - v( {} ) = 0

(1, 2, 3, 0)

v( {0, 1, 2, 3} )- v( {1, 2, 3} ) = 0

v( {1} ) - v( {} ) = 0

(1, 3, 0, 2)

v( {0, 1, 3} )- v( {1, 3} ) = 1

v( {1} ) - v( {} ) = 0

(1, 3, 2, 0)

v( {0, 1, 2, 3} )- v( {1, 2, 3} ) = 0

v( {1} ) - v( {} ) = 0

(2, 0, 1, 3)

v( {0, 2} )- v( {2} ) = 1

v( {0, 1, 2} )- v( {0, 2} ) = 0

(2, 0, 3, 1)

v( {0, 2} )- v( {2} ) = 1

v( {0, 1, 2, 3} )- v( {0, 2, 3} ) = 0

(2, 1, 0, 3)

v( {0, 1, 2} )- v( {1, 2} ) = 1

v( {1, 2} ) - v( {2} ) = 0

(2, 1, 3, 0)

v( {0, 1, 2, 3} )- v( {1, 2, 3} ) = 0

v( {1, 2} ) - v( {2} ) = 0

(2, 3, 0, 1)

v( {0, 2, 3} )- v( {2, 3} ) = 1

v( {0, 1, 2, 3} )- v( {0, 2, 3} ) = 0

(2, 3, 1, 0)

v( {0, 1, 2, 3} )- v( {1, 2, 3} ) = 0

v( {1, 2, 3} )- v( {2, 3} ) = 1

(3, 0, 1, 2)

v( {0, 3} )- v( {3} ) = 1

v( {0, 1, 3} )- v( {0, 3} ) = 0

(3, 0, 2, 1)

v( {0, 3} )- v( {3} ) = 1

v( {0, 1, 2, 3} )- v( {0, 2, 3} ) = 0

(3, 1, 0, 2)

v( {0, 1, 3} )- v( {1, 3} ) = 1

v( {1, 3} ) - v( {3} ) = 0

(3, 1, 2, 0)

v( {0, 1, 2, 3} )- v( {1, 2, 3} ) = 0

v( {1, 3} ) - v( {3} ) = 0

(3, 2, 0, 1)

v( {0, 2, 3} )- v( {2, 3} ) = 1

v( {0, 1, 2, 3} )- v( {0, 2, 3} ) = 0

(3, 2, 1, 0)

v( {0, 1, 2, 3} )- v( {1, 2, 3} ) = 0

v( {1, 2, 3} )- v( {2, 3} ) = 1

Table 3 shows some initially confusing calculations in the last two columns, where each of these columns is defined for a given player. Suppose a player and a permutation are defined. For that permutation, let the set Sπ, i contain those players in the permutation π to the left of the given player i. The difference, in the last two columns, is the following, for i equal to 0 and to 1, respectively:

v(Sπ, i ∪ {i}) - v(Sπ, i)

The Shapley-Shubik power index, for a player, is the ratio of a sum to the number of permutations of players. And that sum is calculated for each player, as the sum over all permutations, of the above difference in the value of the value of the characteristic function.

If I understand correctly, given a permutation, the above difference can only take on values of 0 or 1. And it will only be 1 for one player, where that player determines whether the formation of the coalition in the order given will be a winning coalition. As a consequence, the Shapley-Shubik power index is guaranteed to sum over players to unity. In this case, power is a fixed amount, with each player being measured as having a defined proportion of that power.

5.0 Both Power Indices

The above has stepped through the calculation of two power indices, for all players, in a given game. Table 4 lists their values, as well as a normalization of the Penrose-Banzhauf power index such that the sum of the power, over all players, is unity. (I gather that the Penrose-Banzhauf index and the normalized index do not have the same properties.) As one might expect from the definition of the game, "my aunt" has more power than "me" in this game.

Table 4: The Penrose-Banzhaf and Shapley-Shubik Power Indices

Player

Penrose-Banzhaf Power Index

Shapley-ShubikPower Index

Index

Normalized

0

6/16 = 3/8

6/12 = 1/2

12/24 = 1/2

1

2/16 = 1/8

2/12 = 1/6

4/24 = 1/6

2

2/16 = 1/8

2/12 = 1/6

4/24 = 1/6

3

2/16 = 1/8

2/12 = 1/6

4/24 = 1/6

In many voting games, the normalized Penrose-Banzhauf and Shapley-Shubik power indices are not identical for all players. In fact, suppose the rules for the above variation of Me and my Aunt voting game are varied. Suppose now that four votes - a supermajority - are needed to carry a motion. The normalized Penrose-Banzhaf index for player 0 becomes 1/3, while each of the other players have a normalized Penrose-Banzhaf index of 2/9. Interestingly enough, the Shapley-Shubik indices for the players do not change, if I have calculated correctly. But the values assigned to rows in Table 3 do sometimes vary. Anyways, that one tweak of the rules results in different power indices, depending on which method one adopts. A more interesting example would be one in which the rankings vary among power indices.

Other power indices, albeit less common, do exist. Which one is most widely applicable? I would think that mainstream economists, given game theory and marginalism, would tend to prefer the Shapley-Shubik power index. Felsenthal and Machover (2004) seem to be widely recognized experts on measures of voting power, and they have come to prefer the Penrose-Banzhaf index over the Shapley-Shubik index.

6.0 Where To Go From Here

I have described above a couple of power indices in voting games. As I understand it, many have tried to write down reasonable axioms that characterize power indices. One challenge is to specify a set of axioms such that your preferred power index is the only one that satisfies them. But, as I understand it, some sets of reasonable axioms are open insofar as more than one power index would satisfy them. I seem to recall a theorem that one could create a power index for a reasonable set of axioms such that whichever player you want in a voting game is the most powerful. Apparently, a connection can be drawn between a power index and a voting procedure. And Donald Saari boasts that he could create an apparently fair voting procedure that would result in whatever candidate you like being elected.

I gather that many examples of voting games have been presented in which apparently paradoxical or perverse results arise. And these do not seem to be merely theoretical results. Can I find some such examples? Perhaps, I should look here at some of Daron Acemoglu's work.

I am aware of three types of examples to look for. One is that of a dummy. A dummy is a player that, under the weights and the rule for how many votes are needed for passage, can never be decisive in a coalition. Whether this player drops out or joins a coalition can never change whether or not a resolution is passed, even though the player has a positive weight. A second odd possibility arises as the consequence of adding a new member to the electorate:

"...power of a weighted voting body may increase, rather than decrease, when new members are added to the original body." -- Steven J. Brams and Paul J. Affuso (1976).

A third odd possibility apparently can arise on a council when one district annexes another. Suppose, the district annexing the other consequently increases the weight of its vote accordingly. One might think a greater weight leads to more power. But, in certain cases, the normalized Penrose-Banzhaf index can decrease.

The above calculations for the Penrose-Banzhaf and Shapley-Shubik power indices treat all coalitions or permutations, respectively, as equally likely to arise. Empirically, this does not seem to be true. And this has an impact on how one might measure power. For example, since voting is unweighted on the Supreme Court of the United States, all justices might be thought to be equally powerful. But, because of the formation of well-defined blocks, Anthony Kennedy was often described as being particularly powerful in deciding court decisions, at least when Antonin Scalia was still alive. So empirically, one might include some assessment of the affinities of the players for one another and, thus, some influence on the probabilities of each coalition forming. This will have consequences on the calculation of power indices. But why stop there? In the United States these days, politicians only seem to represent the most wealthy.

Update: This page, from the University of Warwick, has links to utilities for calculating various power indices.